Saving OneToMany , foreign key not set

[Frank VanOor]

I've two tables: TaStock and TaStockPrice. Field tastockid in table TaStockPrice is the foreign key to table TaStock.

@Entity
public class TaStock {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    Integer id

    @OneToMany(mappedBy = "taStock", fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    private List<TaStockPrice> tastockpriceList;

    public void addTaStockPrice(TaStockPrice taStockPrice) {
       if (taStockPrice == null) {
           return;
       }
       taStockPrice.setTaStock(this);
       if (tastockpriceList == null) {
           tastockpriceList = new ArrayList<TaStockPrice>();
           tastockpriceList.add(taStockPrice);
       } else if (!tastockpriceList.contains(taStockPrice)) {
           tastockpriceList.add(taStockPrice);
       }
    }
    ....
}

@Entity
public class TaStockPrice {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    Integer id

    @Column
    private Integer tastockid;


    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "tastockid", nullable = false, updatable = false, insertable = false)
    private TaStock taStock;
    ...
}

persisting taStock with Children

@Test
@Transactional(propagation = Propagation.REQUIRES_NEW)
public void createTaStock() throws Exception {
    TaStock taStock = new TaStock();
             ...

    TaStockPrice taStockPrice = new TaStockPrice();
    taStockPrice.setTaStock(taStock);
             ...

    taStock.addTaStockPrice(taStockPrice);
    taStockService.persist(taStock);
}

I read that when persisting a parent class, hibernate automatically persist the children of that class. But instead, the following exception occurs:

javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: ERROR: null value in column "tastockid" violates not-null constraint

Thanks in advance for help

[Emanuel Kadziela]

You have to generate the entity ids somehow. For example:

    @Id
    @GeneratedValue
    private Long id;

[Frank VanOor]

Thanks for reply,

The Id's were generated by the DB, but I changed the code and added the annotation : @GeneratedValue(strategy = GenerationType.IDENTITY). Still the same problem.

[Alexander Bondarev]

You have to remove the class variable 'tastockid' from TaStockPrice. The foreign key column for the many-to-one relationship is defined in the JoinColumn annotation.
And the column 'tastockid' should be insertable.

[Frank VanOor]

I've commented out the field taStockId in the class TaStockPrice. But still the same exception.

javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: ERROR: null value in column "tastockid" violates not-null constraint


The table in Postgresql looks like this :

CREATE TABLE tastockprice
(
id serial NOT NULL,
tastockid integer NOT NULL,
...
CONSTRAINT tastockprice_pkey PRIMARY KEY (id ),
CONSTRAINT tastockprice_tastockid_fkey FOREIGN KEY (tastockid)
REFERENCES tastock (id) MATCH SIMPLE
ON UPDATE CASCADE ON DELETE CASCADE
)

With my current setup, hibernate doesn't seem to set field tastockid. How Can I make sure that hibernate recognised tastockid as the foreign key, and sets this field when saving TaStock ?

[Alexander Bondarev]

Frank VanOor wrote:javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: ERROR: null value in column "tastockid" violates not-null constraint
...
With my current setup, hibernate doesn't seem to set field tastockid. How Can I make sure that hibernate recognised tastockid as the foreign key, and sets this field when saving TaStock ?

You have this exception because the column 'tastockid' is not included in SQL INSERT statement. The column definition has the element 'insertable = "false"'. Try to change it to 'insertable = "true"'.

[Frank VanOor]

I change the colomn taStockId as follow

@Column(insertable = true)
private Integer tastockid;

Still the same exception :-(


The log of hibernate shows the following output:

Hibernate: insert into TaStock (active, buydate, buysell, selldate, slope, stockid, tapatternid, watch) values (?, ?, ?, ?, ?, ?, ?, ?)
Hibernate: select currval('TaStock_id_seq')
Hibernate: insert into TaStockPrice (date, line, price, quality, tastockid) values (?, ?, ?, ?, ?)

It looks like hibernate doesn't use the Id of TaStock in the insert statement of TaStockPrice.

[Alexander Bondarev]

I meant to change like this:

@Entity  
public class TaStockPrice {  
  
    @Id  
    @GeneratedValue(strategy = GenerationType.IDENTITY)  
    Integer id; 
  
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "tastockid", nullable = false, updatable = false, insertable = true)
    private TaStock taStock;
    ...
}

[Frank VanOor]

Hi Alexander , thank for help.
I changed like you suggested, but have the following exception now.

Caused by: org.hibernate.MappingException: Repeated column in mapping for entity: com.stockdomain.domain.TaStockPrice column: tastockid (should be mapped with insert="false" update="false")
at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:681)
at org.hibernate.mapping.PersistentClass.checkPropertyColumnDuplication(PersistentClass.java:703)
at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:725)
at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:478)
at org.hibernate.mapping.RootClass.validate(RootClass.java:270)
at org.hibernate.cfg.Configuration.validate(Configuration.java:1294)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1736)
at org.hibernate.ejb.EntityManagerFactoryImpl.<init>(EntityManagerFactoryImpl.java:94)
at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:905)

[Alexander Bondarev]

I am guessing this is because you are still having the field 'private Integer tastockid;' in the class TaStockPrice. Remove it from the class definition.

[Frank VanOor]

I removed "private Integer tastockid" from TaStockPrice, and all goes fine !! Rows are inserted in DB.

thank you very much for you patience and help Alexander .

Frank